Let $R$ be a ring not necessarily with unity with an idempotent element $e\neq 0, 1$ induces the (two-sided) Peirce decomposition $R = eRe ⊕ eR(1 − e) ⊕ (1− e)Re ⊕ (1 − e)R(1 − e)$. Here $eRe$ = $\{{ere | r ∈ R\}}$, $eR(1 − e)$, $(1 − e)Re$ and $(1 − e)R(1 − e)$ are abelian subgroups of $R$, where the abbreviated notation $eR(1 − e)$ stands formally for the set $\{{e(r − re) = er − ere | r ∈ R\}}$; and similarly, $(1 − e)Re$ = $\{{re − ere | r ∈ R\}}$, $(1 − e)R(1 − e)$ = $\{{r − er − re + ere | r ∈ R\}}$.
we call this idempotent element $e_1$=$e$ and formally set $e_2$ = $1−e$. Then for $R_{ij}$ = $e_iRe_j$ (i, j = 1, 2) we may write $R$ in its Peirce decomposition $R = R_{11} ⊕ R_{12} ⊕ R_{21} ⊕ R_{22}$. However all these abelian subgroups are also subring but not ideal (proof is easy).
In case of prime ring$R$ , the set $ I = R_{12}R_{21} + R_{12} + R_{21} + R_{21}R_{12} $ is an ideal of the ring $R$.
I want to know the following things,
How we define the $ R_{12}R_{21}$ and $ R_{21}R_{12} $ ? As I found $R_{12}$ and $R_{21}$ are not ideals of the ring R.
How the set $I$ is an ideal in ring R ?
Why do you want to multiply two subsets of a ring? This is not usually done. But anyhow, defining it to be $e_2Re_1Re_2$ makes it an ideal in $R_{22}$, so that seems like a reasonable thing to use.
It's straightforward to verify that when you write $$ I=\begin{bmatrix}eR(1-e)Re & eR(1-e)\\(1-e)Re& (1-e)ReR(1-e) \end{bmatrix} $$
and
$$ R=\begin{bmatrix}eRe & eR(1-e)\\(1-e)Re& (1-e)R(1-e) \end{bmatrix} $$
that $I$ is left and right $R$ absorbing, and also an abelian group (that is the ring structure which is isomorphic to that of $R$.)